# Let r(x)=f(g(h(x))), where h(1)=2, g(2)=3, h'(1)=4, g'(2)=5, and f'(3)=6, how do you find r'(1)?

The value of

The chain rule for two functions is

So it would make sense if

We don't have any concrete functions here to work with, but we do know the values of the functions/derivatives at certain points.

Hopefully this helps!

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To find ( r'(1) ), use the chain rule:

[ r'(x) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) ]

Evaluate each derivative at ( x = 1 ):

[ h(1) = 2, \quad g(2) = 3, \quad h'(1) = 4, \quad g'(2) = 5, \quad f'(3) = 6 ]

[ r'(1) = f'(g(h(1))) \cdot g'(h(1)) \cdot h'(1) ] [ = f'(g(2)) \cdot g'(h(1)) \cdot h'(1) ] [ = f'(3) \cdot g'(2) \cdot h'(1) ] [ = 6 \cdot 5 \cdot 4 ] [ = \boxed{120} ]

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To find ( r'(1) ), we can use the chain rule.

Given:

- ( r(x) = f(g(h(x))) )
- ( h(1) = 2 )
- ( g(2) = 3 )
- ( h'(1) = 4 )
- ( g'(2) = 5 )
- ( f'(3) = 6 )

First, we find ( h'(x) ), ( g'(x) ), and ( f'(x) ) using the given derivatives.

Now, apply the chain rule: [ r'(x) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) ]

Plug in the given values: [ r'(1) = f'(g(h(1))) \cdot g'(h(1)) \cdot h'(1) ]

Substitute the given values: [ r'(1) = f'(g(2)) \cdot g'(2) \cdot h'(1) ]

We're given ( g(2) = 3 ), so: [ r'(1) = f'(3) \cdot g'(2) \cdot h'(1) ]

Substitute the given value for ( f'(3) = 6 ): [ r'(1) = 6 \cdot 5 \cdot 4 ]

Calculate the result: [ r'(1) = 120 ]

Therefore, ( r'(1) = 120 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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